# What is the equation of the line normal to f(x)=6x^2 -x - 9  at x=1?

Mar 10, 2016

$y = - \frac{x}{11} - \frac{43}{11}$

#### Explanation:

$f \left(x\right) = 6 {x}^{2} - x - 9$

We can find the gradient of this function $m$ at $x = 1$ by finding the 1st derivative, then we can get the gradient of the normal $m '$ using the fact that $m . m ' = - 1$.

$f ' \left(x\right) = 12 x - 1 = m$

So at $x = 1 :$

$m = \left(12 \times 1\right) - 1 = 11$

$\therefore m ' = - \frac{1}{11}$

$f \left(1\right) = 6 - 1 - 9 = - 4$. This is the $y$ value at $x = 1$

The equation of the normal line is of the form:

$y = m ' x + c '$

$\therefore - 4 = - \frac{1}{11} + c '$

$\therefore c ' = \frac{1}{11} - 4 = \frac{1}{11} - \frac{44}{11} = - \frac{43}{11}$

So the equation of the normal$\Rightarrow$

$y = - \frac{x}{11} - \frac{43}{11}$

The situation looks like this:

graph{(-x/11-43/11-y)(6x^2-x-9-y)=0 [-20, 20, -10, 10]}